Question:
Why zinc does not exhibit variable oxidation states?
Why zinc does not exhibit variable oxidation states?
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Zinc has a stable $3d^{10}$ configuration.
Updated On: Oct 1, 2026
- Completely filled \(4s\) subshell
- Completely filled \(3d\) subshell
- Incomplete \(3d\) subshell
- Incomplete \(4s\) subshell
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The Correct Option is B
Solution and Explanation
Step 1: Write the configuration
Zn is \([\text{Ar}]3d^{10}4s^2\).
Step 2: Reason
The \(3d\) subshell is completely filled and very stable. Zinc loses only its two \(4s\) electrons to form Zn\(^{2+}\) and does not disturb the \(3d^{10}\) set. So it shows only the \(+2\) state.
The \(4s\) subshell is also full, but the full \(3d\) shell is the reason given. Option (B).
Final Answer:
Zinc has a completely filled \(3d\) subshell, option (B). \[ \boxed{\text{(B)}} \]
Zn is \([\text{Ar}]3d^{10}4s^2\).
Step 2: Reason
The \(3d\) subshell is completely filled and very stable. Zinc loses only its two \(4s\) electrons to form Zn\(^{2+}\) and does not disturb the \(3d^{10}\) set. So it shows only the \(+2\) state.
The \(4s\) subshell is also full, but the full \(3d\) shell is the reason given. Option (B).
Final Answer:
Zinc has a completely filled \(3d\) subshell, option (B). \[ \boxed{\text{(B)}} \]
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